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  1. Practice
  2. Algorithms
  3. Strings
  4. Anagram
  5. Discussions

Anagram

  • tao_zhang + 8 comments

    My java solution:

    import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-->0){
            String str = sc.next();
            int len = str.length(), count = 0;
            if (len%2!=0){
                System.out.println(-1);
                continue;
            }
            String s1 = str.substring(0,len/2);
            String s2 = str.substring(len/2,len);
            char[] s1Chars = s1.toCharArray();
            for (char c : s1Chars){
                int index = s2.indexOf(c);
                if (index == -1){
                    count++;
                } else {
                    s2 = s2.substring(0,index)+s2.substring(index+1);
                }
            }
            System.out.println(count);
        }
    }
}
    • mithildeeva + 0 comments

      awesome! such simple logic. Thankyou!

    • adikiit123 + 0 comments

      Will it work with the string bbxxxaxb ?

    • humoyun_ahmad + 0 comments

      core of this solution is indexOf() method

    • tvucic1 + 0 comments

      Beautiful!!

    • Gurparshad + 0 comments

      amazing , very helpful

    • varlyanijatin88 + 1 comment

      Solution is Amazing. But can you tell me why did you added s2.substring(index+1); Thanks in advance.

      • shweta78 + 0 comments

        to remove duplicacy... e.g. in let s1='abbc' and s2='fabc'. Since count of b isn't same in s1 and s2, not using s2 = s2.substring(0,index)+s2.substring(index+1); would consider b to be present ignoring the count. Expl: s2.substring(0,index) would calc the substring from beginning to pos index-1(including) and s2.substring(index+1) means starting from index+1 position thereby omitting the character at pos index.

    • ritikagarwal + 1 comment

      what does while (t-->0) means in above programs

      • prageeth_maha + 0 comments

        Super late to the game but, it means while( t-- > 0) {...} It does decrement. Same can be like while(t++

    • kris_tobiasson + 1 comment

      Here's my JS solution:

          if(s.length%2 !== 0){ return -1; }
    let sub = s.substring(0, s.length * 0.5).split(``);

    for(let i = sub.length; i < s.length; i++){
        let id = sub.indexOf(s.charAt(i));
        
        if(id !== -1){
            sub.splice(id,1);
        }
    }
    return sub.length;
      • saxenarajat26 + 1 comment

        "this is working but time limit exceeds can u help me with this!!"

        function anagram(s) { var l=s.length,c=0,c1,c2 if(l%2==0){ c1=s.slice(0,l/2).split("") c2=s.slice(l/2).split("") }else{ return "-1" } var l2=c1.length for(var i=0;i

        • kris_tobiasson + 1 comment

          i would love to help you, but your code is incomplete and it isnt formatted correctly. If you supply the whole solution properly formatted, I would be happy to help.

          By the way, with comments there is no code block button, but you can put the code within three backticks at the beginning and three backticks at the end of the code block to have it properly formatted.

          Cheers

          • saxenarajat26 + 1 comment

            ` 'use strict';

            const fs = require('fs');

            process.stdin.resume(); process.stdin.setEncoding('utf-8');

            let inputString = ''; let currentLine = 0;

            process.stdin.on('data', inputStdin => { inputString += inputStdin; });

            process.stdin.on('end', _ => { inputString = inputString.replace(/\s*Undefined control sequence \n/, ''));

            main();

            });

            function readLine() { return inputString[currentLine++]; }

            // Complete the anagram function below. function anagram(s) { var l=s.length,c=0,c1,c2 if(l%2==0){ c1=s.slice(0,l/2).split("") c2=s.slice(l/2).split("") var l2=c1.length for(var i=0;i

            return c }

            function main() { const ws = fs.createWriteStream(process.env.OUTPUT_PATH);

            const q = parseInt(readLine(), 10);

for (let qItr = 0; qItr < q; qItr++) {
    const s = readLine();

    let result = anagram(s);

    ws.write(result + "\n");
}
    ```

ws.end();

            }

            • kris_tobiasson + 2 comments

              All I need is the full anagram() function.. Anyhow, if you take a look in the comment you just sent, you will see that in the middle of the anagram() function, it cuts off at a for loop statment for(var i=0;i at the i and then goes straight to return c; , so you are missing some code. Just supply the whole anagram function, no need for the other code.

              • saxenarajat26 + 0 comments
                [deleted]
              • kris_tobiasson + 1 comment

                I saw the code in your PM. The performance hit was in your for loop. You were checking to see if one substring had the other substring's character twice (once with includes() and once with indexOf()). Both those functions will search for the given character in a string/array, but includes() will return true or false and indexOf() will return the index of the character or -1 if it doesn't contain the character. So, if you only use indexOf() and check if the return value is === -1 instead of searching through the string twice. Below is an edited version of the code you sent me and you will see what i mean:

                    let l=s.length,c=0,c1,c2;

    if(l%2==0){
        c1=s.slice(0,l/2).split("");
        c2=s.slice(l/2).split("");
        let l2=c1.length;

        for(let i=0;i<l2;i++){
            let id = c2.indexOf(c1[i]);
            if(id !== -1){
                c2[id]=" ";
            }
            else{ c+=1; }
        }
    }
    else{ return "-1"; }
return c;
                • saxenarajat26 + 0 comments

                  thanks i got it !! that includes was unnecessary taking so much time!!